Let A = begin{bmatrix} 1 & -4 & 7 0 & 3 & -5 | vedclass

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(B) The augmented matrix is $[A: B] = \begin{bmatrix} 1 & -4 & 7 & : & a \ 0 & 3 & -5 & : & -b \ -2 & 5 & -9 & : & -c \end{bmatrix}$.Applying row op

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$a = \frac{b + c}{2}$

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(B) The augmented matrix is $[A: B] = \begin{bmatrix} 1 & -4 & 7 & : & a \ 0 & 3 & -5 & : & -b \ -2 & 5 & -9 & : & -c \end{bmatrix}$.Applying row operation $R_3 ightarrow R_3 + 2R_1$:$[A: B] = \begin{bmatrix} 1 & -4 & 7 & : & a \ 0 & 3 & -5 & : & -b \ 0 & -3 & 5 & : & -c + 2a \end{bmatrix}$.Applying row operation $R_3 ightarrow R_3 + R_2$:$[A: B] = \begin{bmatrix} 1 & -4 & 7 & : & a \ 0 & 3 & -5 & : & -b \ 0 & 0 & 0 & : & -c + 2a - b \end{bmatrix}$.Since the rank of $A$ is $ho(A) = 2$, for the rank of $[A: B]$ to be equal to the rank of $A$, the last row of the augmented matrix must be zero.Therefore, $-c + 2a - b = 0$, which implies $2a - b - c = 0$ or $2a = b + c$.

Let $A = \begin{bmatrix} 1 & -4 & 7 \\ 0 & 3 & -5 \\ -2 & 5 & -9 \end{bmatrix}$ and $B = \begin{bmatrix} a \\ -b \\ -c \end{bmatrix}$. If $A$ and $[A: B]$ have the same rank, then:

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